3.1 Groups of automorphisms of fields

Consider fields $E\supset F$ . An $F$ -isomorphism $E\rightarrow E$ is called an $F$ -automorphism of $E$ . The $F$ -automorphisms of $E$ form a group, which we denote $\operatorname{Aut}(E/F)$ .

Example 3.1

(a) There are two obvious automorphisms of $\mathbb{C}$ , namely, the identity map and complex conjugation. We’ll see later (LABEL:te16) that by using the Axiom of Choice we can construct uncountably many more.

(b) Let $E=\mathbb{C}(X)$ . A $\mathbb{C}{}$ -automorphism of $E{}$ sends $X$ to another generator of $E$ over $\mathbb{C}{}$ . It follows from (LABEL:te17a) below that these are exactly the elements $\frac{aX+b}{cX+d}$ , $ad-bc\neq 0$ . Therefore $\operatorname{Aut}(E/\mathbb{C})$ consists of the maps $f(X)\mapsto f\left(\frac{aX+b}{cX+d}\right)$ , $ad-bc\neq 0$ , and so

\operatorname{Aut}(E/\mathbb{C})\simeq\operatorname{PGL}_{2}(\mathbb{C}),

the group of invertible $2\times 2$ matrices with complex coefficients modulo its centre. Analysts will note that this is the same as the automorphism group of the Riemann sphere. Here is the explanation. The field $E$ of meromorphic functions on the Riemann sphere $\mathbb{P}_{\mathbb{C}}^{1}$ consists of the rational functions in $z$ , i.e., $E=\mathbb{C}(z)\simeq\mathbb{C}(X)$ , and the natural map $\operatorname{Aut}(\mathbb{P}_{\mathbb{C}}^{1})\rightarrow\operatorname{Aut}(E% /\mathbb{C})$ is an isomorphism.

\operatorname{PGL}_{3}(\mathbb{C})=\operatorname{Aut}(\mathbb{P}_{\mathbb{C}}^% {2})\hookrightarrow\operatorname{Aut}(\mathbb{C}(X_{1},X_{2})/\mathbb{C}),

but this is very far from being surjective. When there are even more variables $X$ , the group is not known. The group $\operatorname{Aut}(\mathbb{C}(X_{1},\ldots,X_{n})/\mathbb{C})$ is the group of birational automorphisms of projective $n$ -space $\mathbb{P}_{\mathbb{C}}^{n}$ , and is called the Cremona group. Its study is part of algebraic geometry (Wikipedia: Cremona group).

In this section, we’ll be concerned with the groups $\operatorname{Aut}(E/F)$ when $E$ is a finite extension of $F$ .

Proposition 3.2

Let $E$ be a splitting field of a separable polynomial $f$ in $F[X]$ ; then $\operatorname{Aut}(E/F)$ has order $[E\colon F].$

Proof.

As $f$ is separable, it has $\deg f$ distinct roots in $E$ . Therefore Proposition 2.7 shows that the number of $F$ -homomorphisms $E\rightarrow E$ is $[E\colon F]$ . Because $E$ is finite over $F$ , all such homomorphisms are isomorphisms. _□

Example 3.3

Consider a simple extension $E=F[\alpha]$ , and let $f$ be a polynomial in $F[X]$ having $\alpha$ as a root. If $\alpha$ is the only root of $f$ in $E$ , then $\operatorname{Aut}(E/F)=1$ by (2.1b). For example, if $\sqrt[3]{2}$ is the real cube root of $2$ , then $\operatorname{Aut}(\mathbb{Q}[\sqrt[3]{2}]/\mathbb{Q})=1$ . As another example, let $F$ be a field of characteristic $p\neq 0$ , and let $a$ be an element of $F$ that is not a $p$ th power. Let $E$ be a splitting field of $f=X^{p}-a.$ Then $f$ has only one root in $E$ (see 2.11), and so $\operatorname{Aut}(E/F)=1$ .

These examples show that, in the statement of the proposition, is necessary that $E$ be a splitting field of a separable polynomial.

When $G$ is a group of automorphisms of a field $E$ , we set

E^{G}=\operatorname{Inv}(G)=\{\alpha\in E\mid\sigma\alpha=\alpha\text{, all }% \sigma\in G\}.

It is a subfield of $E$ , called the subfield of $G$ -invariants of $E$ or the fixed field of $G$ .

In this section, we’ll show that, when $E$ is the splitting field of a separable polynomial in $F[X]$ and $G=\operatorname{Aut}(E/F)$ , then the maps

M\mapsto\operatorname{Aut}(E/M),\quad H\mapsto\operatorname{Inv}(H)

give a one-to-one correspondence between the set of intermediate fields $M$ , $F\subset M\subset E$ , and the set of subgroups $H$ of $G$ .

Theorem 3.4 (E. Artin)

Let $G$ be a finite group of automorphisms of a field $E$ , then

[E\colon E^{G}]\leq(G\colon 1).

Proof.

Let $F=E^{G}$ , and let $G=\{\sigma_{1},\ldots,\sigma_{m}\}$ with $\sigma_{1}$ the identity map. It suffices to show that every set $\{\alpha_{1},\ldots,\alpha_{n}\}$ of elements of $E$ with $n>m$ is linearly dependent over $F$ . For such a set, consider the system of linear equations

\displaystyle\sigma_{1}(\alpha_{1})X_{1}+\cdots+\sigma_{1}(\alpha_{n})X_{n}

\displaystyle=0

equation (6) (6)

\displaystyle\vdots\qquad\qquad

\displaystyle\sigma_{m}(\alpha_{1})X_{1}+\cdots+\sigma_{m}(\alpha_{n})X_{n}

\displaystyle=0

with coefficients in $E$ . There are $m$ equations and $n>m$ unknowns, and hence there are nontrivial solutions in $E$ . We choose one $(c_{1},\ldots,c_{n})$ having the fewest possible nonzero elements. After renumbering the $\alpha_{i}$ , we may suppose that $c_{1}\neq 0$ , and then, after multiplying by a scalar, that $c_{1}\in F$ . With these normalizations, we’ll show that all $c_{i}\in F$ , and so the first equation

\alpha_{1}c_{1}+\cdots+\alpha_{n}c_{n}=0

(recall that $\sigma_{1}$ is the identity map) is a linear relation on the $\alpha_{i}$ .

If not all $c_{i}$ are in $F$ , then $\sigma_{k}(c_{i})\neq c_{i}$ for some $k\neq 1$ and $i\neq 1$ . On applying $\sigma_{k}$ to the system of linear equations

\displaystyle\sigma_{1}(\alpha_{1})c_{1}+\cdots+\sigma_{1}(\alpha_{n})c_{n}

\displaystyle=0

\displaystyle\vdots\qquad\qquad

\displaystyle\sigma_{m}(\alpha_{1})c_{1}+\cdots+\sigma_{m}(\alpha_{n})c_{n}

\displaystyle=0

and using that $\{\sigma_{k}\sigma_{1},\ldots,\sigma_{k}\sigma_{m}\}=\{\sigma_{1},\ldots,% \sigma_{m}\}$ ( $\sigma_{k}$ merely permutes the $\sigma_{i}$ ), we find that

(c_{1},\sigma_{k}(c_{2}),\ldots,\sigma_{k}(c_{i}),\ldots)

is also a solution to the system of equations (6). On subtracting it from the first solution, we obtain a solution $(0,\ldots,c_{i}-\sigma_{k}(c_{i}),\ldots)$ , which is nonzero (look at the $i$ th entry), but has more zeros than the first solution (look at the first entry) — contradiction. _□

Corollary 3.5

Let $G$ be a finite group of automorphisms of a field $E$ ; then

G=\operatorname{Aut}(E/E^{G}).

Proof.

As $G\subset\operatorname{Aut}(E/E^{G})$ , we have inequalities

[E\colon E^{G}]\overset{\text{\ref{ft10}}}{\leq}(G\colon 1)\leq(\operatorname{% Aut}(E/E^{G})\colon 1)\overset{\text{\ref{sf8}a}}{\leq}[E\colon E^{G}].

All the inequalities must be equalities, and so $G=\operatorname{Aut}(E/E^{G}).$ _□